How do i know what resistor value, that's attached to a BLDC motor, i should use in order to get a torque of 100 lb-ft put on my actuator?

by John David Heinzmann (edited) A little more context here would help. As in, what are you trying to do? Sounds like maybe you want to use the motor/actuator as a brake? Or a damper? Is that what you mean by running the actuator in "reverse"? You want the actuator to absorb mechanical power rather than supply mechanical power? By actuator, it seems you might mean a gearbox?

When using a brushed DC motor and SI units, the torque constant and the back EMF constant are numerically equal (they are both expressing the same physical phenomenon of interaction of an electrical conductor with a magnetic field). Thus, a motor with Ke = 25 VDC / krpm, in SI units Ke = 0.239 V/(rad/s). That's the open-circuit voltage generated at the terminals as a function of shaft speed. When using it as a motor, the torque produced will be 0.239 Nm/A = Kt, the torque constant. (By the way, 1 V/(rad/s) = 1 Nm/A = 1 weber! The weber is the unit of magnetic flux.)

Unfortunately, it is a bit more complicated for a brushless DC motor, which is actually an AC motor, and a 3-phase motor at that. If you drive the motor mechanically it will act as a generator and produce a 3-phase back-emf voltage wave form, in your case with 38.95 Vpk/krpm or 0.372 Vpk/(rad/s). I can show you the math if you need me to, but I long ago reverted to using MathCAD (free from PTC) to do these calculations because it completely carries the units through the analysis. Indeed, you can rectify that voltage and if you apply a resistor across that rectified voltage current will flow through the motor and the resistor and cause a braking torque to appear at the shaft of the motor (in the direction that opposes the direction of rotation regardless of direction of rotation). However, it is not a matter of simply applying ohms law to the rectified DC voltage to select the resistor.

1. Depending on the efficiency of the gearbox, the friction in a 60:1 gear ratio when back driven can appear as a significant drag torque and absorb quite a bit of the mechanical power.

2. The three-phase configuration of the motor will require a 3-phase rectifier bridge (six diodes in a full-bridge configuration like in the alternator of a car).

3. The forward voltage drop of the diodes in the bridge will have a non-linear effect on the load curve, especially at low speeds when Vemf is low and the diode drop voltage is significant by comparison.

4. The resistance of the windings will absorb some of the power

5. The inductance of the windings will limit the current that can flow at higher frequencies (motor speeds) and thus the amount of braking torque the motor will provide.

For all these reasons, it may be easier for you to determine the needed resistance experimentally. At 50 rpm and 2.5 ft-lb, you are only talking 17.7 W of mechanical power which, because of losses along the way means your resistor will be absorbing less than that, maybe only 10 or 12 W. That is not a very big resistor so buying a handful of power resistors or even a variable power resistor in the neighborhood of the resistance you needed may suffice.

At 3000 rpm, the back emf will be (314 rad/s)*0.372 Vpk/(rad/s) = 117 Vpk. Since we know the power that will need to be absorbed is less than 17.7 W, the resistor current will have to be less than 17.7 W/117 VDC = 0.15 ADC. Using ohm's law, the resistance will have to be more than 117 V / 0.15 A = 770 Ω. In reality, if the power that needs to be absorbed is more like 12 W, the resistance will be more like 1.1 kΩ.

With all that in mind, you might want to buy a 1.5 kΩ adjustable power resistor and lower the resistance until you get the torque you want. To be safe, it should probably be at least 30 W rated resistor. I see on Digi-key that they only have 1 kΩ and 500 Ω resistors in stock so maybe buy one of these https://www.digikey.com/product-detail/en/vishay-huntington-electric-inc/AVT05006E1K000KE/AVT50-1-0K-ND/257637 and one of these https://www.digikey.com/products/en?keywords=AVT05006E500R0KE and put them in series? You get the idea.

By the way, contrary to your original assumption, the higher the resistance the lower the torque. A short circuit (0 Ω) would give you the highest torque because it would cause the highest current.

Your motor rated for 479 oz-in = 2.59 lb-ft @ 10 Arms (current limit)

Thus 100 lb-ft needs 40x more powerful torque just to hold with no acceleration !!

No Go. !

It seems like you're trying to use a BLDC motor as a passive, constant toque load by shorting the leads together.

This is not possible. When shorting the leads of a BLDC motor it behaves variable torque load. The following equations relate motor speed and torque to voltage and current: $$ V = K_edot{theta} $$ $$tau = K_{tau}I $$
Substuting, the equation to determine the motor torque when driven as a generator with the leads shorted together is as follows:

$$ tau = K_{tau}frac{K_e dot{theta}}{R}$$

As you can see, the motor load torque is directly proportional to the rotor velocity. Assuming your actuator could maintain a controlled 3000rpm, the motor in theory would only produce ~48ft-lb of torque maximum (with the leads all shorted together). Of course, according to the datasheet, the motor is only rated for 10A of current, which means that the motor would be slag long before it got there.

As a side note, 100lb-ft is quite a bit of torque. This motor when driven actively is only rated to produce ~2.4ft-lb. I'm wondering if maybe you made a mistake with your units somewhere.